\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x=m\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^22x=m\)
\(\Leftrightarrow1-\dfrac{1}{4}\left(1-cos4x\right)=m\)
\(\Leftrightarrow3+cos4x=4m\)
\(\Leftrightarrow cos4x=4m-3\)
a.
Với \(m=1\) pt trở thành:
\(cos4x=1\Leftrightarrow4x=k2\pi\)
\(\Leftrightarrow x=\dfrac{k\pi}{2}\)
b. Do \(-1\le cos4x\le1\) nên pt vô nghiệm khi:
\(\left[{}\begin{matrix}4m-3>1\\4m-3< -1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m>1\\m< \dfrac{1}{2}\end{matrix}\right.\)
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